Question 1189249
<font face="Times New Roman" size="+2">


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 19x\ +\ 3y\ =\ 224]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{224\ -\ 3y}{19}]


*[tex \Large 19] goes into *[tex \Large 224\ \ ]*[tex \Large 11] times with a remainder of *[tex \Large 15], so in order for *[tex \Large x] and *[tex \Large y] to be integers, *[tex \Large 19x\ =\ 19*11\ =\ 209] and *[tex \Large 3y\ =\ 224\ -\ 209\ =\ 15] so


*[tex \Large x\ =\ 11] and *[tex \Large y\ =\ 5] and *[tex \Large x\ +\ y\ =\ 16]

																
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
*[illustration darwinfish.jpg]

From <https://www.algebra.com/cgi-bin/upload-illustration.mpl> 
I > Ø
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
{{n}\choose{r}}
</font>